Math Problem Statement

why does n+1/n^2 converge and show simplification of the expression in detail

Solution

To understand why the series n=1n+1n2\sum_{n=1}^{\infty} \frac{n+1}{n^2} converges, we need to analyze and simplify the expression n+1n2\frac{n+1}{n^2} and then examine its behavior as nn becomes large.

1. Simplifying the Expression

Let's start by simplifying the expression n+1n2\frac{n+1}{n^2}:

n+1n2=nn2+1n2\frac{n+1}{n^2} = \frac{n}{n^2} + \frac{1}{n^2}

Now, simplify each term individually:

nn2=1nand1n2 is already simplified.\frac{n}{n^2} = \frac{1}{n} \quad \text{and} \quad \frac{1}{n^2} \text{ is already simplified.}

So, the expression becomes:

n+1n2=1n+1n2\frac{n+1}{n^2} = \frac{1}{n} + \frac{1}{n^2}

2. Analyzing the Series

The series n=1n+1n2\sum_{n=1}^{\infty} \frac{n+1}{n^2} can now be written as:

n=1(1n+1n2)=n=11n+n=11n2\sum_{n=1}^{\infty} \left(\frac{1}{n} + \frac{1}{n^2}\right) = \sum_{n=1}^{\infty} \frac{1}{n} + \sum_{n=1}^{\infty} \frac{1}{n^2}

This splits the original series into two separate series:

  1. The Harmonic Series n=11n\sum_{n=1}^{\infty} \frac{1}{n}:

    • This series is known to diverge.
  2. A p-Series n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2}:

    • This is a p-series with p=2p = 2, which is greater than 1. Such a series converges.

3. Combined Behavior of the Series

Even though the harmonic series diverges, the term 1n2\frac{1}{n^2} decreases much faster than 1n\frac{1}{n} as nn increases. The rapid decay of 1n2\frac{1}{n^2} is sufficient to make the combined series n=1(1n+1n2)\sum_{n=1}^{\infty} \left(\frac{1}{n} + \frac{1}{n^2}\right) converge.

4. Formal Convergence Argument

To confirm convergence, consider the limit comparison test:

  • We already know 1n2\sum \frac{1}{n^2} converges. Let's compare n+1n2\frac{n+1}{n^2} with 2n2\frac{2}{n^2} as nn grows large:

limnn+1n21n2=limn(n+11)=limn(n+1)=\lim_{n \to \infty} \frac{\frac{n+1}{n^2}}{\frac{1}{n^2}} = \lim_{n \to \infty} \left(\frac{n+1}{1}\right) = \lim_{n \to \infty} \left(n + 1\right) = \infty

Instead, let's compare it directly to the harmonic series, where:

n+1n22nn2=2n\frac{n+1}{n^2} \leq \frac{2n}{n^2} = \frac{2}{n}

Thus, we have:

n+1n22n2\sum \frac{n+1}{n^2} \sim \sum \frac{2}{n^2}

Since 2n2\sum \frac{2}{n^2} converges, the original series n+1n2\sum \frac{n+1}{n^2} must also converge.

5. Conclusion

The series n=1n+1n2\sum_{n=1}^{\infty} \frac{n+1}{n^2} converges because, although it contains a divergent harmonic series term 1n\frac{1}{n}, the addition of the convergent term 1n2\frac{1}{n^2} ensures the overall series decreases rapidly enough to converge.


Would you like more details or have any questions?

Here are some related questions:

  1. What is the limit comparison test and how does it work for series convergence?
  2. How does the behavior of 1n\frac{1}{n} compare to 1n2\frac{1}{n^2} as nn increases?
  3. Why is the harmonic series divergent even though it decreases?
  4. What are other examples of series that converge due to the rapid decrease of their terms?
  5. How do we determine the convergence of a series involving a combination of divergent and convergent terms?

Tip: Simplifying the general term of a series and comparing it to a known convergent or divergent series is often a useful strategy for determining convergence.

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Math Problem Analysis

Mathematical Concepts

Series convergence
Harmonic series
p-Series

Formulas

Sum of series

Theorems

Limit comparison test

Suitable Grade Level

Advanced undergraduate level